H3PO4 Stepwise Neutralization

  • Thread starter lkh1986
  • Start date
In summary: Whereas in the H2SO4 – NaOH case, if the Ka3 is very small, then the H+ from the H3PO4 will also contribute to the pH. However, in the Na2HPO4 – NaOH case, since the Ka3 is much larger, then only the H+ from the NaOH will contribute to the pH.
  • #1
lkh1986
99
0
For the triprotic acid, phosphoric acid, H3PO4, there are Ka1, Ka2,
and Ka3. And we know from the book that the value of Ka3 is too
small, so we neglect the [H+] contribution from it.

Normally, we encounter questions involving pH calculations of
neutralization of a diprotic acid and a base, for example, H2SO4 and
NaOH. Let say there are 5 mmol of H2SO4 and 10 mmol of NaOH.

The stepwise reactions are:

H2SO4 + NaOH ---> NaHSO4 + H2O
---5-----------10------------0---------------
---0-----------5--------------5---------------
NaHSO4 + NaOH ---> Na2SO4 + H2O
----5-----------5-------------0---------------
----0-----------0--------------5--------------

However, if the reaction is between H3PO4 and NaOH, let say there are
5 mmol H3PO4 and 15 mmol NaOH.

The stepwise reactions are:

H3PO4 + NaOH ---> NaH2PO4 + H2O
----5---------15------------0-----------------
----0---------10------------5-----------------

NaH2PO4 + NaOH ---> Na2HPO4 + H2O
----5------------10--------------0---------------
----0-------------5--------------5---------------

Na2HPO4 + NaOH ---> Na3PO4 + H2O
----5------------5--------------0--------------
----0------------0--------------5--------------

So, are the 3 equations above correct? I mean, can we use the 3
equations above for calculating the pH value, as what we would do for
the case of the neutralization involving a diprotic acid and NaOH?

If so, why do we neglect the [H+] from H3PO4?

Or is it that we ONLY neglect the [H+] from Ka3 when we calculate the
pH of a solution containing ONLY H3PO4, but we should
somehow "include" and "consider" the Ka3 when we wish to find the pH
value of a solution involving neutralization of H3PO4 and NaOH, since
this neutralization reaction is based on the stoichiometric ratio.

Thanks.
 
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  • #2
Or is it that we ONLY neglect the [H+] from Ka3 when we calculate the
pH of a solution containing ONLY H3PO4, but we should
somehow "include" and "consider" the Ka3 when we wish to find the pH
value of a solution involving neutralization of H3PO4 and NaOH, since
this neutralization reaction is based on the stoichiometric ratio.

You are right.

You can prove that H3PO4 third dissocitation is negligible as long as H3PO4 is the only source of hydronium ions in solution. ( either generally or u can take a numerical example and try it out yourself . )

But notice that the reactions u posted :
H3PO4 + NaOH ---> NaH2PO4 + H2O
NaH2PO4 + NaOH ---> Na2HPO4 + H2O
Na2HPO4 + NaOH ---> Na3PO4 + H2O

are not entirely quantitative. that is , the advancement decreases from one reaction to the other. For each rxn , k = ka/kw . Notice the 3rd rxn is much weaker than the 1st reaction , which is pretty much 'complete'. You can find the ka's and calculate the K of each reaction to convince urself. Finding the pH here is different from that of sulfuric acid.
 
  • #3


I can confirm that the three equations provided for the neutralization of H3PO4 and NaOH are correct. However, when calculating the pH of a solution involving this neutralization reaction, it is important to consider the contribution of all three Ka values.

The reason we typically neglect the contribution of Ka3 is due to its small value, which means it has a minimal effect on the overall pH of the solution. However, it is still important to consider it in the calculations, especially when dealing with large amounts of H3PO4 and NaOH.

Additionally, when neutralizing a solution of only H3PO4, we can neglect the contribution of Ka3 because there is no base present to react with it. But in the case of neutralization with NaOH, we must consider all three Ka values to accurately calculate the pH of the solution.

In summary, while Ka3 may be negligible in some cases, it should still be considered in the calculations for neutralization of H3PO4 and NaOH to ensure accurate results.
 

1. What is H3PO4 and why is it important?

H3PO4, also known as phosphoric acid, is a type of acid commonly used in various industrial processes and in the production of fertilizers. It is important because it is a key component in the phosphorus cycle, which is essential for the growth of plants and the production of food.

2. What is stepwise neutralization and how does it work with H3PO4?

Stepwise neutralization is a chemical process in which an acid is gradually neutralized by adding a base. In the case of H3PO4, it has three acidic protons that can be neutralized one at a time. This means that H3PO4 can undergo stepwise neutralization three times, resulting in different products depending on the number of protons neutralized.

3. How is H3PO4 stepwise neutralization used in the production of fertilizers?

H3PO4 stepwise neutralization is used in the production of fertilizers to create different types of phosphates that are essential for plant growth. By neutralizing one or more of the acidic protons in H3PO4, different types of phosphates with varying solubility and availability to plants can be produced.

4. What are the products of H3PO4 stepwise neutralization?

The products of H3PO4 stepwise neutralization depend on the number of protons neutralized. If one proton is neutralized, the product is monobasic phosphate (H2PO4-). If two protons are neutralized, the product is dibasic phosphate (HPO42-). And if all three protons are neutralized, the product is tribasic phosphate (PO43-).

5. How does H3PO4 stepwise neutralization affect the pH of a solution?

H3PO4 stepwise neutralization can affect the pH of a solution depending on the number of protons neutralized. Each neutralization reaction results in the formation of a conjugate base, which can either be acidic or basic. Therefore, the pH of the solution will change as the neutralization reactions occur.

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